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Finding the Angle -- Cannon fired from a patrol boat aimed at the top of a building

  • #1

Homework Statement



A patrol boat off a coast is ordered to destroy a rocket launcher that sits on the roof of 40-meter tall building. The horizontal distance between the boat and the rocket launcher is 1500 meters. The muzzle velocity of the patrol boat’s stabilized 30-mm cannon is 141 meters/second. What are the two angles (relative to the horizontal) at which the patrol boat’s gun can be aimed in order to destroy the rocket launcher? Express your answer in degrees to a precision of three significant figures. [You may ignore air resistance. Assume an acceleration of gravity of precisely 9.800 m/s2. Your answers must be accurate to within 0.3 degrees in order to receive credit. You must also show your work.]
Vo= 141 m/s
X= 1500 m
Yf = 40 m
Yo = 0
t = ?
Theta = ?

Homework Equations


Yf=Yo + Voy(t) - .5(9.8)(tsquared)

The Attempt at a Solution


One angle seems to be 23.4 degrees but I cant understand how you would find the other angle. I divided the initial velocity by the x distance to find 10.34 seconds. Plugged that into the Yf=Yo + Voy(t) - .5(9.8)(tsquared) equation with Yf being 40 meters, Yo being zero and Voy being (Vo)(sin)(theta) and then solving for theta. I checked it and it is accurate within a meter to the final position of y being 40 meters so the math is accurate even if my theory may be incorrect. However, I dont see any strategy for finding the second angle.
 
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Answers and Replies

  • #2
haruspex
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divided the initial velocity by the x distance to find 10.34
I assume you mean you divided the horizontal distance by the velocity, but that is not the horizontal velocity.
 
  • #3
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Your one equation has two unknowns: v0y and t. That will not be enough. You need another equation. You should probably consider the motion in x.
 
  • #4
Anyone have any ideas on what to do here?
 
  • #5
haruspex
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Anyone have any ideas on what to do here?
Let the angle be θ. What is the horizontal velocity? What is the time to the target?
 
  • #6
Let the angle be θ. What is the horizontal velocity? What is the time to the target?
Yes, Im aware that the angle is theta. There is no other information provided, You must deduce the angle from ONLY those three variables - Vo is 141 m/s. Final y position I assume is 40 meters and the x distance is 1500 meters. Ive taken the derivative for time to be t = X / (Vo) (Cos) (Theta) and then solving for theta which Ive worked is the logical means to the solution. I plug my derivative for T into the equation Yf = Yi + gt - 4.9 t squared but unfortunately my math skill are too broken to be able to work the math out sufficiently. Ive attempted three times and checked it against the equation and it does not work. Basically, Im just asking someone to confirm my theory and solve the math for my equation which becomes, with all variables, 40 = 0 + (141) [( 1500/ (141) (Cos) (Theta)] - 4.9 ([( 1500/ (141) (Cos) (Theta)] Squared and then solving for Theta. My math skills are not good enough to solve this equation.
 
  • #7
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There's a square root in there, so you need to consider both the positive and negative roots.
 
  • #8
898
382
You are very close. Using the x equation to get t in terms of v0 and cos(theta) and substituting into the y equation is one way to do it. You might find it easier to solve the y equation for t and substitute it into x. As you correctly noted, the rest is just solving the equation. We’ll talk about that in a second.

However, first, I do see one mistake. You wrote

Yf = Yi + gt - 4.9 t squared

Why g t? g isn’t the initial velocity in the y dimension (hint hint)

Now, about solving the equation. You will find you have a result which includes both sines and cosines, and which can’t be easily solved for theta. Note that your problem statement indicates that the answer must be within 0.3 degrees. This is a good indication that the problem requires approximation or a numerical solution. You can either approximate the trig functions, or find the two solutions numerically
 
  • #9
haruspex
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40 = 0 + (141) [( 1500/ (141) (Cos) (Theta)] - 4.9 ([( 1500/ (141) (Cos) (Theta)] Squared
Not quite. Please explain all the terms in this part: (141) [( 1500/ (141) (Cos) (Theta)]
 

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